Problem: Simplify and expand the following expression: $ \dfrac{1}{2y - 20}- \dfrac{3}{5y + 40}+ \dfrac{4y}{y^2 - 2y - 80} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{1}{2y - 20} = \dfrac{1}{2(y - 10)}$ We can factor a $5$ out of denominator in the second term: $ \dfrac{3}{5y + 40} = \dfrac{3}{5(y + 8)}$ We can factor the quadratic in the third term: $ \dfrac{4y}{y^2 - 2y - 80} = \dfrac{4y}{(y - 10)(y + 8)}$ Now we have: $ \dfrac{1}{2(y - 10)}- \dfrac{3}{5(y + 8)}+ \dfrac{4y}{(y - 10)(y + 8)} $ The least common multiple of the denominators is: $ 10(y - 10)(y + 8)$ In order to get the first term over $10(y - 10)(y + 8)$ , multiply by $\dfrac{5(y + 8)}{5(y + 8)}$ $ \dfrac{1}{2(y - 10)} \times \dfrac{5(y + 8)}{5(y + 8)} = \dfrac{5(y + 8)}{10(y - 10)(y + 8)} $ In order to get the second term over $10(y - 10)(y + 8)$ , multiply by $\dfrac{2(y - 10)}{2(y - 10)}$ $ \dfrac{3}{5(y + 8)} \times \dfrac{2(y - 10)}{2(y - 10)} = \dfrac{6(y - 10)}{10(y - 10)(y + 8)} $ In order to get the third term over $10(y - 10)(y + 8)$ , multiply by $\dfrac{10}{10}$ $ \dfrac{4y}{(y - 10)(y + 8)} \times \dfrac{10}{10} = \dfrac{40y}{10(y - 10)(y + 8)} $ Now we have: $ \dfrac{5(y + 8)}{10(y - 10)(y + 8)} - \dfrac{6(y - 10)}{10(y - 10)(y + 8)} + \dfrac{40y}{10(y - 10)(y + 8)} $ $ = \dfrac{ 5(y + 8) - 6(y - 10) + 40y} {10(y - 10)(y + 8)} $ Expand: $ = \dfrac{5y + 40 - 6y + 60 + 40y}{10y^2 - 20y - 800} $ $ = \dfrac{39y + 100}{10y^2 - 20y - 800}$